Question

If \[{N_2}{H_4}\] is oxidised to\[{N_2}\], what would be the equivalent weight of \[{N_2}{H_4}\]?

Answer 1

Hint: Equivalent weight of the substance is the mass of that particular substance which will replace \[1g\] of hydrogen, \[8g\]of oxygen or \[35.5g\]of chlorine. Equivalent weight can be determined by knowing the oxidation states of a particular atom or an element in a compound.

Complete answer:
To solve this question first we need to know about the \[{N_2}{H_4}\]and oxidation states of nitrogen in hydrazine and in nitrogen gas.
So, \[{N_2}{H_4}\] is commonly known as hydrazine, and is a colourless dense liquid with a strong smell. Hydrazine is a strong reducing agent so let’s see its oxidation state.
For \[{N_2}{H_4}\]
$2x + (1 \times 4) = 0$
$x = - 2 $
So, the oxidation state of N in \[{N_2}{H_4}\] is \[ - 2\] .
And we already know, by convention, the oxidation state of nitrogen gas \[{N_2}\] is\[0\]. Thus, we can see the change in the oxidation state from \[ - 2\] to \[0\] be\[2\].
Since, in hydrazine, \[{N_2}{H_4}\] we are having two nitrogen atoms \[{N_2}\] , so we will multiply it by 2
So our n = \[4\]
Now let’s recall the formula of equivalent weight,
\[Equivalent{\text{ }}weight = \dfrac{{gram{\text{ atomic weight }}}}{{valence}}\]
Since, the atomic weight of hydrazine \[{N_2}{H_4}\]is \[32\]
So, putting the value of atomic weight and n in formula, we get,
\[Equivalent{\text{ }}weight = \dfrac{{32{\text{ }}}}{4} = 8\]
Therefore, the answer to our question is 8 i.e. the equivalent weight of \[{N_2}{H_4}\] , if it gets oxidized to \[{N_2}\] , will be \[8\] .

Note:
Hydrazine has many applications like it is used as a rocket fuel propellant in combination with dinitrogen tetroxide . Hydrazine is really toxic as it gets absorbed on skin .hydrazine is best known for spreading invisible fire as it has no flames .